# Limits Problem worked 2 ways gives 2 different answers

I am confused about finding a limit.

$$\lim_{x\to\infty}\sqrt{x^2+4x+1}-x$$

I know the answer is $2$ and I can get to that answer by first multiplying by the complex conjugate and then dividing by the highest power of $x$ (which is $x$). However, when I try to do this without multiplying by the complex conjugate I get a different answer as follows:

Divide both pieces of the expression by $\frac{1}{x}$. This $\frac{1}{x}$ factor becomes $\frac{1}{x^2}$ when it moves inside the square root. So, I would get $$\sqrt{\frac{1}{x^2}(x^2+4x+1)}-\frac{1}{x} x$$

This would lead to $$\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}-1$$

taking the limit as $x$ goes to infinity would then give $\sqrt{1} -1 = 0 \neq 2$. Not sure where I am going wrong. Any suggestions?

• The complex conjugate, or just the conjugate? May 9 '14 at 17:09
• You divide something that is supposedly 2 by $x$. $x\to\infty$. Two divided by something going to infinity approaches $0$. $0$ is the result you get. Why are you surprised? ;-) May 20 '14 at 9:33

Sure, you may be multiplying both pieces of the expresison by $\frac 1 x$, but then that's not the same limit. The limit you have to calculate is:

$$\lim_{x\rightarrow+\infty}x\frac 1 x\left(\sqrt{x^2+4x+1}-x\right)$$

If you multiply the thing inside the parentheses by $\frac 1 x$ you also have to multiply it by $x$ or you're not taking the same limit. And then:

$$\lim_{x\rightarrow+\infty}x\left(\sqrt{1+\frac 4 x+\frac 1 {x^2}}-1\right)$$

And that limit is of the form $\infty*0$ which can probably be solved using l'Hôpital.

How come you are just dividing by $x$?

You should be multiplying and dividing by $x$. You are just dividing your expression which is indeed $2$ by a large number, it has to become zero.

Better way : $$\sqrt{(x+2)^2-3}-x$$

As $x\to \infty$, $\sqrt{(x+2)^2}-x=2$

• Could you formalize the argument, i'd like to use it on tests : ) Dec 15 '19 at 22:15

The reason the second method won't work is because what you actually have is $$\lim_{x\to\infty}\frac{\sqrt{x^2+4x+1}-x}{1}.$$If you divide the numerator by $x$, then you have to divide the denominator by $x$ as well. That is, you are essentially multiplying the fraction by $1$, because $\dfrac{\frac{1}{x}}{\frac{1}{x}}=1.$ As I'm sure you are aware, multiplying any expression by $1$ does not change the value of the expression. If you multiply the numerator by $\dfrac{1}{x},$ and not the denominator also, then you are changing the problem.

You can't just divide by $x$. Why should the limit remain the same? It's not a quotient where you could divide both numerator and denominator by the same value. One easy way to find the limit in such cases is

$$\lim_{x\to\infty}\sqrt{x^2+4x+1}-x=\lim_{x\to\infty}\frac{x^2+4x+1-x^2}{\sqrt{x^2+4x+1}+x}=\lim_{x\to\infty}\frac{4+\frac{1}{x}}{\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}+1}=2$$